Question

Let \( (\mathbb{R}^2, d_1) \) and \( (\mathbb{R}^2, d_2) \) be two metric spaces with \[ d_1\left( (x_1, x_2), (y_1, y_2) \right) = |x_1 - y_1| + |x_2 - y_2| \] \[ {and} \quad d_2\left( (x_1, x_2), (y_1, y_2) \right) = \frac{d_1\left( (x_1, x_2), (y_1, y_2) \right)}{1 + d_1\left( (x_1, x_2), (y_1, y_2) \right)}. \] If the open ball centered at \( (0,0) \) with radius \( \frac{1}{7} \) in \( (\mathbb{R}^2, d_1) \) is equal to the open ball centered at \( (0,0) \) with radius \( \frac{1}{\alpha} \) in \( (\mathbb{R}^2, d_2) \), then the value of \( \alpha \) is (in integer).

Hint:

In metric spaces, the relationship between the radii of open balls can be used to compute the corresponding values in different metrics. Here, we used the formula for \( d_2 \) to find the relation between the radii.

Answer 1

Correct Answer: 8

To solve the problem, we examine the properties and relationships of the specified metric spaces. We have two metrics, \(d_1\) and \(d_2\), defined on \(\mathbb{R}^2\). The metric \(d_1\) is the Manhattan distance, while \(d_2\) is a normalized version of \(d_1\).

Step 1: Understand the metrics

The metric \(d_1((x_1,x_2),(y_1,y_2))\) is defined as:

\[ d_1((x_1,x_2),(y_1,y_2)) = |x_1 - y_1| + |x_2 - y_2| \] 

For \(d_2\), the formula is:

\[ d_2((x_1,x_2),(y_1,y_2)) = \frac{d_1((x_1,x_2),(y_1,y_2))}{1 + d_1((x_1,x_2),(y_1,y_2))} \]

Step 2: Set up the conditions

We are given that the open ball centered at \((0,0)\) with radius \(\frac{1}{7}\) in \((\mathbb{R}^2,d_1)\) is equivalent to the open ball centered at \((0,0)\) with radius \(\frac{1}{\alpha}\) in \((\mathbb{R}^2,d_2)\).

Step 3: Calculate the equivalence of radii

The radius of the open ball in the \(d_1\) metric is \(\frac{1}{7}\), so we have:

\[ d_1((x_1,x_2),(0,0)) = |x_1| + |x_2| < \frac{1}{7} \]

For metric \(d_2\), the ball's radius is \(\frac{1}{\alpha}\):

\[ \frac{|x_1| + |x_2|}{1 + |x_1| + |x_2|} < \frac{1}{\alpha} \]

Equating the two conditions, set \(\frac{1}{\alpha}\) equal to the transformed radius of \(\frac{1}{7}\):

\[ \frac{r_1}{1 + r_1} = \frac{1}{7} \]

Substitute \(r_1 = \frac{1}{7}\):

\[ \frac{\frac{1}{7}}{1 + \frac{1}{7}} = \frac{\frac{1}{7}}{\frac{8}{7}} = \frac{1}{8} \]

We learn the radius \(r_2 = \frac{1}{\alpha} = \frac{1}{8}\), so \(\alpha = 8\).

Step 4: Validate against the range

The calculated value \(\alpha = 8\) is within the provided range [8,8], confirming its validity.

Thus, the integer value of \(\alpha\) is \(8\).